# Compound interest

A user deposits $3.000$ € in a saving account with an interest per year of $6\%$ but instead of withdrawing the interests he keeps it in the deposit, so that it increases the inverted quantity and generates, in this way, new interest.

How much money will be in the account after $5$ years?

If the liquidation period is annual, that is to say, the bank deposits the interest in the account when the year is over, an option is to calculate the obtained simple interest every year and to accumulate in one lump sum at the end of every year.

We will call the interest generated in the first year $I_1$:

$$I_1=C\cdot r \cdot t=3.000\cdot 0,06\cdot 1=180€$$

If we add up these $180$ € of interest obtained in the first year to the initial $3.000$ €, at the beginning of the second year there will be $3.180$ € in the account, which will be the initial quantity to calculate the simple interest of the second period:

$$I_2=3.180\cdot 0,06\cdot 1=190,8€$$

So, at the end of the second year, there will be $3.180+190,8 = 3.370,8$ €, accumulated, which will be the initial capital of the third year:

$$I_3=3.370,8\cdot 0,06\cdot 1=202,25€$$

Accumulated capital $=3370,8+202,25=3573,05€$

The interest in the fourth year will be:

$$I_4=3.573,05\cdot 0,06\cdot 1=214,38€$$

Accumulated capital $=3573,05+214,38=3787,43€$

And, finally, at the end of the five years the account will have reached:

$$I_5=3.787,43\cdot 0,06\cdot 1=227,25€$$

Total amount accumulated $=3787,43+227,25=4014,68$€

A faster way of realizing these operations is by means of the relation that defines the compound interest:

$$C_f=C\cdot (1+r)^t$$

where $C_f$ is the final capital, $C$ is te initial capital, $r$ the yield (interest in an amount of unitary fraction), and $t$ is the time.

Once we have applied the relation to the example, we get:

$$C_f=3.000\cdot (1+0,06)^5=3.000\cdot (1,06)^5=3.000\cdot 1,3382=4.014,6€$$

In this case, applying the relation, it has been immediate because the yield and the time are both annual. Nevertheless, this case is not usually the most frequent.

The banks and the savings banks handle several liquidation periods: monthly, quarterly, half-yearly, annual, etc. And obviously, it is not the same that the interests increase the total amount invested on the following month than after one year. In order to verify the formula of the compound interest it is necessary to express the yield and the time according to the period of liquidation that the bank uses.

For example:

A client deposits $10.000$ € in an account with annual interest of $4,5\%$. If the interests are accumulating in the account every trimester: what quantity of money will be there after $5$ years? What will the interest generated in the above mentioned period be?

The first thing that we have to do is to realize that the liquidation period is quarterly and that, therefore, the yield and the time will have to be expressed according to this fact, that is to say, it will be necessary to find out the quarterly yield and the number of trimesters that there are in $5$ years.

To find the quarterly yield, that is called $r(t)$,we take the expression between parentheses of the formula of the compound interest and this condition has to be satisfied:

$$(1+r_t)^4=(1+r)^1$$

That is, the yield of four trimesters is equal to the yield of one year, since $1$ year has $4$ trimesters.

Knowing by the statement that $r = 0,045$, we only need to isolate $r(t)$ to find out this information. To do so, we have firstly take the exponent and put it on the other side of the equality:

$$1 + r_t =(1+r)^{\frac{1}{4}} \Rightarrow r_t=(1+r)^{\frac{1}{4}} -1$$

The quarterly yield is calculated with the information that we already have:

$$r_t=(1+0,045)^{\frac{1}{4}} -1 = (1,045)^{\frac{1}{4}} -1=1,011-1=0,011$$

Now we only have to know how many quarterly liquidation periods there are in $5$ years:

If $1$ year has $4$ trimesters, $5$ will have:

$$4\cdot 5=20 \ \mbox{trimesters}$$

With the information obtained it is already possible to apply the formula of the compound interest to calculate the money accumulated at the end of the period:

$$C_f=C \cdot (1+r_t)^t$$

$$C_f=10.000 \cdot (1+0,011)^20=10.000\cdot (1,011)^20=$$ $$=10.000\cdot 1,245 = 12.450€$$

With that, the first question of the exercise has been solved. To find the second we just have to subtract the final quantity and the initial, since the obtained interests will be the difference between the two:

$$12.450-10.000=2.450€$$

It is necessary to notice the importance of computing the yield depending on the period of liquidation as it been done in the exercise, thereby avoiding surprises.

For example, if a bank offers a deposit at $12\%$ per year with a period of half-yearly liquidation we might think that the half-yearly yield will be:

$$\mbox{If} \ r_a=0,12 \Rightarrow r_s=\dfrac{0,12}{2}=0,06$$

Since in one year there are $2$ half-yearly liquidation periods, it seems logical to think that if the annual interest is $12\%$, the half-yearly one will be $6\%$, but in fact it is not completely like that.

To demonstrate we just need to calculate the half-yearly yield as it has appeared in the previous exercise:

$$(1+r_s)^2=(1+r)^1 \Rightarrow 1+r_s=(1+r)^{\frac{1}{2}} \Rightarrow r_s=(1+r)^{\frac{1}{2}} -1$$

$$r_s=(1+0,12)^{\frac{1}{2}} -1 = (1,12)^{\frac{1}{2}} - 1=1,0583-1=0,0583$$

Namely, the half-yearly interest is not $6\%$, as we might think, but $5,83\%$.

One last example:

A savings bank granted a loan of $120.000$ € to a company. After the agreed period of $8$ years, the savings bank received a quantity of $20.599,13$ € as interest, which was paid annually. What was the interest on the above mentioned loan?

It is a question of isolating the yield of the relation of the compound interest and converting it into a percentage:

$$C_f=C(1+r)^t \Rightarrow \dfrac{C_f}{C}=(1+r)^t \Rightarrow \Big( \dfrac{C_f}{C}\Big)^{\frac{1}{t}}=1+r \Rightarrow$$

$$\Rightarrow r=\Big( \dfrac{C_f}{C}\Big)^{\frac{1}{t}} -1$$

The final quantity will be the initial amount plus the generated interests:

$$C_f=120.000+20.599,13=140.599,13$$

Now we can calculate the yield with the rest of the information in the statement:

$$r=\Big( \dfrac{140.599,13}{120.000} \Big) ^{\frac{1}{8}} -1 = 1,172^{\frac{1}{8}} -1 = 1,02-1=0,02$$

So, the interest of the loan was $2\%$.